Alright, let's tackle this problem using the properties of the normal distribution.
1. Identify Given Information:
- Average (mean) cost ([tex]\(\mu\)[/tex]): [tex]$22
- Standard deviation (\(\sigma\)): $[/tex]3
- Threshold cost: [tex]$25
2. Determine What to Find:
- The probability that the cost will exceed $[/tex]25.
3. Calculate the Z-Score:
The Z-score helps us understand how many standard deviations away our threshold value ([tex]$25) is from the mean. The Z-score formula is:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) is the threshold cost ($[/tex]25),
- [tex]\(\mu\)[/tex] is the mean ([tex]$22),
- \(\sigma\) is the standard deviation ($[/tex]3).
So, plugging in the values:
[tex]\[
Z = \frac{25 - 22}{3} = \frac{3}{3} = 1
\][/tex]
4. Using the Normal Distribution:
- The Z-score of 1 tells us that the cost of [tex]$25 is 1 standard deviation above the mean.
- To find the probability of the cost being more than $[/tex]25, we need to look at the area to the right of Z = 1 in the normal distribution curve.
5. Empirical Rule and the Normal Distribution Table:
- The empirical rule states that approximately 68% of data falls within 1 standard deviation (σ) of the mean (µ). This leaves about 32% outside this range, divided between the two tails of the normal distribution.
- Specifically, about 16% (or 0.15865525393145707 to be exact) lies above 1 standard deviation from the mean in the right tail.
6. Conclusion:
The probability that at any given theater, the cost will be more than [tex]$25 for two people to go to a movie is approximately:
\[
\text{Probability} \approx 0.15865525393145707
\]
Converting this to a percentage:
\[
\text{Probability} \approx 15.87\%
\]
Therefore, the probability that at any given theater, the cost will be more than $[/tex]25 for two people to go to a movie is approximately 15.87%.
